Behavioral Resistor Laplace Expression

QSPICE
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This post is to ask about/clarify the Laplace expression convention for the behavioral resistor.

I tried to implement a frequency-dependent resistor (resistance decreases as frequency increases) with a behavioral resistor and a Laplace expression. I first tried “Laplace = 1 / (abs(s) / (2pi))” but this caused the resistance to increase with frequency. The expression “Laplace = abs(s) / (2pi)” achieved what I wanted, but it seems backwards to me. I assumed the Laplace expression defined the impedance of the behavioral resistor. Would need the “s” term in the denominator to decrease the resistance with frequency.

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Does the Laplace expression set the impedance or admittance? Is this discussed in the documentation? Apologies if I just missed something simple. Thanks!

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Not sure if the Laplace resistor is officially a thing. This question was discussed in the LTspice forum years ago : LTspice@groups.io | Laplace resistor

Here is an explanation of why Laplace formula is inverted with G-source : Tutorials

Are you simulating skin effect? For your information, laplace resistor will not work in time domain simulation.

B Source - Behavioral R with Laplace.qsch (2.5 KB)

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Thanks for the confirmation and pointing me to the LTspice forum discussion/the tutorial! I hadn’t seen those links. Very much appreciated.

I am simulating dielectric losses for a capacitive sensing application. Luckily only need .ac analysis for now :slight_smile:

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